Equations, Inequalities, and Systems

Algebraic Reasoning at Certification Level

Linear and absolute-value equations, inequalities, and systems anchor the algebra portion of Praxis 5165. With only 66 questions in 180 minutes, multi-step items must be executed cleanly. Many stems embed these skills inside word problems or student-response analysis.

Linear Equations

Use inverse operations. Keep equations balanced: whatever you do to one side, do to the other.

Worked Example 1

Solve 3(2x − 4) = 5x + 6.

6x − 12 = 5x + 6 → x = 18. Check: 3(36 − 4) = 96; 5(18) + 6 = 96.

Worked Example 2 — Variables on both sides

7 − 2(3 − x) = 4x + 1 → 7 − 6 + 2x = 4x + 1 → 2x − 4x = 1 − 1 → −2x = 0 → x = 0.

Absolute Value Equations

If |A| = k and k ≥ 0, then A = k or A = −k. If k < 0, no real solution.

Worked Example 3 — |2x − 5| = 7

2x − 5 = 7 → x = 6.

2x − 5 = −7 → x = −1.

Solution set: {−1, 6}.

Worked Example 4 — |x + 3| = −2

Absolute value cannot equal a negative number in reals → no solution.

Worked Example 5 — |3x − 1| = |x + 7|

Either 3x − 1 = x + 7 → x = 4, or 3x − 1 = −(x + 7) → 4x = −6 → x = −3/2. Check both in the original.

Linear Inequalities

Solve like equations except flip the inequality when multiplying or dividing by a negative.

Worked Example 6

Solve −4x + 3 > 11.

−4x > 8 → x < −2 (inequality reversed).

Graph: open circle at −2, shade left.

Compound inequalities such as −3 < 2x + 1 ≤ 9 solve in two parts on all sections: −4 < 2x and 2x ≤ 8 → −2 < x ≤ 4.

Systems of Linear Equations

Two equations, two unknowns. Solution is an ordered pair satisfying both.

Worked Example 7 — Elimination

2x + y = 11 x − y = 1

Add equations: 3x = 12 → x = 4. Then y = 3. Check in both originals.

Worked Example 8 — Substitution

y = 2x − 1 and 3x + y = 14.

Substitute: 3x + 2x − 1 = 14 → 5x = 15 → x = 3, y = 5.

Worked Example 9 — Scaling elimination

4x + 3y = 18 and 2x − y = 4.

Multiply second equation by 3: 6x − 3y = 12. Add to first: 10x = 30 → x = 3, y = 2.

Special System Cases

• No solution: parallel lines (same slope, different intercept)—inconsistent system.
• Infinitely many solutions: same line—dependent system.

Example: x + y = 5 and 2x + 2y = 11 are parallel → no solution.

Teaching scenarios may show a student adding equations incorrectly; re-run the algebra before endorsing a method.

Inequality Systems (Preview)

A system like y > x + 1 and y ≤ −2x + 8 describes a feasible region. Shade intersections on the plane; boundary lines are dashed when strict inequality holds.

Strategy Checklist

• Distribute before collecting like terms.
• For absolute value, always consider two cases (or compare two absolute values with two cases).
• When eliminating, multiply one or both equations to align coefficients.
• After solving a system, substitute into both equations—students often satisfy only one.
• On teaching items, a valid method that reaches the correct solution should not be marked wrong solely for being longer than elimination.

Clearing Fractions in Linear Equations

Worked Example 10

Solve (x/3) + (x/2) = 5. LCD = 6: 2x + 3x = 30 → x = 6.

Multiply by the LCD early to avoid fraction errors—especially under time pressure on a 66-question exam.

Literal Equations

Solve A = 2πr + πr^2 for r in terms of A when r > 0. This is less common than linear literal equations such as d = rt → t = d/r.

For y = mx + b, solving for x gives x = (y − b)/m—useful when interpreting slope as rate of change.

Three-Variable Preview

Praxis 5165 focuses on two-variable systems, but recognize inconsistent and dependent classifications quickly. If two equations simplify to the same line, the system has infinitely many solutions; if parallel, none.

Graphical Interpretation

The solution to a system is the intersection point. If lines are parallel, there is no intersection in the plane. A teaching item might show a graph where the student picked the intersection of two lines that are not both constraints—re-read which lines represent the equations given.